AMC10 2024 B

AMC10 2024 B · Q14

AMC10 2024 B · Q14. It mainly tests Probability (basic), Area & perimeter.

A dartboard is the region $B$ in the coordinate plane consisting of points $(x, y)$ such that $|x| + |y| \le 8$. A target $T$ is the region where $(x^2 + y^2 - 25)^2 \le 49$. A dart is thrown and lands at a random point in B. The probability that the dart lands in $T$ can be expressed as $\frac{m}{n} \cdot \pi$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$?

飞镖盘是坐标平面中区域 $B$，由满足 $|x| + |y| \le 8$ 的点 $(x, y)$ 组成。靶心 $T$ 是区域 $(x^2 + y^2 - 25)^2 \le 49$。飞镖随机落在 $B$ 中的一点。飞镖落在 $T$ 中的概率可表示为 $\frac{m}{n} \cdot \pi$，其中 $m$ 和 $n$ 互质正整数。求 $m + n$？

(A) 39 39

(B) 71 71

(D) 75 75

(E) 135 \qquad 135 \qquad

Answer

Correct choice: (B)

正确答案：（B）

Solution

Inequalities of the form $|x|+|y| \le 8$ are well-known and correspond to a square in space with centre at origin and vertices at $(8, 0)$, $(-8, 0)$, $(0, 8)$, $(0, -8)$. The diagonal length of this square is clearly $16$, so it has an area of \[\frac{1}{2} \cdot 16 \cdot 16 = 128\] Now, \[(x^2 + y^2 - 25)^2 \le 49\] Converting to polar form, \[r^2 - 25 \le 7 \implies r \le \sqrt{32},\] and \[r^2 - 25 \ge -7\implies r\ge \sqrt{18}.\] The intersection of these inequalities is the circular region $T$ for which every circle in $T$ has a radius between $\sqrt{18}$ and $\sqrt{32}$, inclusive. The area of such a region is thus $\pi(32-18)=14\pi.$ The requested probability is therefore $\frac{14\pi}{128} = \frac{7\pi}{64},$ yielding $(m,n)=(7,64).$ We have $m+n=7+64=\boxed{\textbf{(B)}\ 71}.$

形如 $|x|+|y| \le 8$ 的不等式对应以原点为中心、顶点在 $(8, 0)$、$(-8, 0)$、$(0, 8)$、$(0, -8)$ 的正方形。该正方形对角线长显然为 $16$，面积为 \[\frac{1}{2} \cdot 16 \cdot 16 = 128\] 现在， \[(x^2 + y^2 - 25)^2 \le 49\] 转为极坐标形式， \[r^2 - 25 \le 7 \implies r \le \sqrt{32},\] \[r^2 - 25 \ge -7\implies r\ge \sqrt{18}.\] 这些不等式的交集是环形区域 $T$，半径介于 $\sqrt{18}$ 和 $\sqrt{32}$（含端点）。该区域面积为 $\pi(32-18)=14\pi$。故概率为 $\frac{14\pi}{128} = \frac{7\pi}{64}$，$(m,n)=(7,64)$。因此 $m+n=7+64=\boxed{\textbf{(B)}\ 71}$。

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